时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input
Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:
ID K ID[1] ID[2] ... ID[K]where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID‘s of its children. For the sake of simplicity, let us fix the root ID to be 01.
Output
For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output "0 1" in a line.
1 #include <iostream>
2
3 using namespace std;
4
5
6
7 struct fun
8
9 {
10
11 int father;
12
13 bool ifdo;
14
15 int h;
16
17 };
18
19
20
21 fun F[101];
22
23
24
25 int main()
26
27 {
28
29
30
31 int n;
32
33 while(cin>>n)
34
35 {
36
37 int i;
38
39 for(i=1;i<=n;i++)
40
41 {
42
43 F[i].ifdo=true;
44
45 F[i].h=1;
46
47 F[i].father=-1;
48
49 }
50
51
52
53 int m;
54
55 cin>>m;
56
57 int high=0;
58
59 if(m==0) cout<<1<<endl;
60
61 else
62
63 {
64
65
66
67 int name1,name2,num;
68
69 while(m--)
70
71 {
72
73 cin>>name1>>num;
74
75
76
77 F[name1].ifdo=false;
78
79 for(i=0;i<num;i++)
80
81 {
82
83 cin>>name2;
84
85 F[name2].father=name1;
86
87 }
88
89
90
91 }
92
93
94
95
96
97 /*这题目陷阱在这，每行第一个ID并不是
98
99 从小到大按循序排列的，所以不能在上个循环中直接求h
100
101 需要记录父节点 在以下循环中求h；
102
103 */
104
105
106
107
108
109 for(i=2;i<=n;i++)
110
111 {
112
113 F[i].h=F[F[i].father].h+1;
114
115 if(F[i].h>high) high=F[i].h;
116
117 }
118
119
120
121
122
123 int j;
124
125 bool fir=true;
126
127 for(i=1;i<=high;i++)
128
129 {
130
131 int sum=0;
132
133 for(j=1;j<=n;j++)
134
135 {
136
137 if(F[j].ifdo&&F[j].h==i) sum++;
138
139 }
140
141 if(fir) cout<<sum;
142
143 else cout<<" "<<sum;
144
145 fir=false;
146
147 }
148
149 cout<<endl;
150
151 }
152
153
154
155
156
157 }
158
159
160
161
162
163 return 0;
164
165 }
166
167
168